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If $A$ is an $n$-by-$n$ matrix with complex entries, (i.e., $A\in M_n(\mathbb{C})$,) $A$ must have $n$ eigenvalues, counting algebraic multiples. But it is not always true that $A$ has $n$ linearly independent eigenvectors. So, what necessary and sufficient condition may be add, to ensure that $A$ has $n$ linearly independent eigenvectors?

Of course, the simpler the better.

I have another related question:

I can't figure out why, intuitively, that algebraic multiple doesn't mean more than one linearly independent eigenvectors. I mean in my tuition, a multiple appear because there is a subspace with dimension>1 being scaled "evenly" in every direction. If the multiple is 2, how come I may fail to find 2 linearly independent vectors in this subspace?

xzhu
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2 Answers2

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Iff $\,A\,$ is diagonalizable iff the minimal polynomial of A is a product of different linear factors...

There you have two (equivalent, of course) conditions

DonAntonio
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  • Thank, you. You're right but these are the answers I know (or can find in textbook.) What I really want to know is that, WHAT exactly, is the intrinsic differences, as linear transformations, between a diagonalizable matrix and a non-diagonalizable matrix. – xzhu Sep 10 '12 at 03:20
  • I realize this doesn't best describe what I'm wondering, so I might as well start a new question. – xzhu Sep 10 '12 at 03:23
  • @Voldemort: the intrinsic difference is the presence of nontrivial Jordan blocks. See http://en.wikipedia.org/wiki/Jordan_normal_form . Another relevant keyword here is "semisimple module over $\mathbb{C}[x]$" (see http://en.wikipedia.org/wiki/Semisimple_module). – Qiaochu Yuan Sep 10 '12 at 04:24
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For example, gcd$(f(X), f'(X)) = 1$, where $f(X)$ is the minimal polynomial of $A$. gcd$(f(X), f'(X))$ can be calculated by the Euclid's algorithm.

Makoto Kato
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